Answer:
vaporization
Explanation:
The molar enthapy of _vaporization______ is the heat required to vaporize one mole of a liquid”
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The correct answer is 0.15.
We are aware that there is 0.05 mol of an unidentified hydrocarbon we will refer to as "X" and that its burning produces 6.6 g of carbon dioxide and 3.6 g of water.
These quantities might be converted to moles by applying the following formula:
amount= mass/ relative atomic mass
Thus, the following equation may be written for H2O: moles = 3.6 / 18 = 0.2 and for CO2: moles = 6.6 / 44 = 0.15.
0.05X + x'O2 = 0.15CO2 + 0.2H2O
This may be made simpler by dividing through by 0.05 (this step is likely to be the most helpful to you), resulting in:
1 x + x O2 = 3 co2 + 4 H2O
The hydrocarbon must have been the source of all the carbon in the carbon dioxide and all the hydrogen in the water.
Accordingly, 4 x 2 = 8 moles of H and 3 x 1 = 3 moles of C.
There are 3/1 = 3 Cs and 8/1 = 8 Hs in one X molecule.
This clearly identifies C3H8 or propane as the hydrocarbon X (dividing by 1 seems unnecessary, but it illustrates the process to use if there were more than one mol of X in the first equation).
To learn more about number of moles of carbon dioxide refer the link:
brainly.com/question/12723070
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Answer:
Noble gas, any of the seven chemical elements that make up Group 18 (VIIIa) of the periodic table. The elements are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og)
Explanation:
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